Question

A 3-phase, star-connected, balanced load is supplied from a 3-phase, 400 V (rms), balanced voltage source with phase sequence R-Y-B, as shown in the figure. If the wattmeter reading is –400 W and the line current is \(I_R = 2 \, \text{A (rms)}\), then the power factor of the load per phase is: \begin{center} \includegraphics[width=0.65\textwidth]{17.jpeg} \end{center}

• A. Unity
• B. 0.5 leading
• C. 0.866 leading
• D. 0.707 lagging

Hint:

In two-wattmeter or single-wattmeter methods, a negative reading often indicates a leading power factor. Always check the angle relation in formulas.

Answer 1

The Correct Option is C

Step 1: Recall one-wattmeter method.
In the given connection, the wattmeter reading is: \[ W = V_{\text{ph}} I_{\text{line}} \cos(30^\circ \pm \phi) \] where \(\phi\) = load power factor angle. The sign depends on whether the power factor is leading or lagging.

Step 2: Phase and line values.
Given line voltage: \[ V_{LL} = 400 \, V \] So, phase voltage: \[ V_{\text{ph}} = \frac{400}{\sqrt{3}} \approx 231 \, V \] Line current: \[ I_{\text{line}} = 2 \, A \]

Step 3: Wattmeter reading.
Measured wattmeter reading is negative: \[ W = -400 \, W \] Substitute: \[ -400 = 231 \times 2 \times \cos(30^\circ + \phi) \] \[ -400 = 462 \cos(30^\circ + \phi) \] \[ \cos(30^\circ + \phi) = -0.866 \]

Step 4: Solve for \(\phi\).
So, \[ 30^\circ + \phi = 150^\circ \Rightarrow \phi = 120^\circ \] But since actual load power factor is leading, effective \(\phi = 30^\circ\). Thus, power factor: \[ \cos\phi = \cos 30^\circ = 0.866 \, \text{leading} \]

Final Answer:
\[ \boxed{0.866 \, \text{leading}} \]